Question

$2A+3B\to 4C+5D$
Starting with 7 moles of $A$, 6 moles of $B$. Find the moles of $C$ and $D$ formed respectively.

• A. $8\text{mol},10\text{mol}$
• B. $14\text{mol},17.5\text{mol}$
• C. $9\text{mol},11.25\text{mol}$
• D. $12\text{mol},15\text{mol}$

Answer 1

The correct option is A

$8\text{mol},10\text{mol}$

$2A+3B\to 4C+5D$
Dividing the given moles of reactant by the respective stoichiometric coefficient of that reactant we get,
A $=\frac{7}{2}=3.5$ and $B=\frac{6}{3}=2$
Therefore, $B$ is the limiting reagent.
Hence using the law of stoichiometry the moles of C formed $=\frac{(4\times 6)}{3}=8\text{mol}$
Similarly, moles of D formed $=\frac{(5\times 6)}{3}=10\text{mol}$